47
votes
Why isn't the Weierstrass function $\sum_{n=0}^\infty a^n \cos(b^n\pi x)$ differentiable?
Nothing is preventing us from taking the derivative of any finite partial sum of this series. This is a trigonometric polynomial and it has derivatives of all orders.
However, this infinite sum ...
- 14.9k
31
votes
Does there exist a function which converts exponentiation into addition?
As Stinking Bishop mentions in the comments, if $f(a)+f(b)=f(a^b)$, then it follows (by interchanging $a$ and $b$) that $f(b)+f(a)=f(b^a)$. Hence, $f(a^b)=f(b^a)$ for all $a,b\in\mathbb R^+$. Setting $...
- 14.8k
25
votes
Why isn't the Weierstrass function $\sum_{n=0}^\infty a^n \cos(b^n\pi x)$ differentiable?
As other answers have pointed out, while the partial sums of the function converge, the partial sums of the derivatives do not. To see this, I just did a quick calculation. Let $f_m(x)$ be the partial ...
- 1,081
21
votes
Accepted
How does this definition of a computable function work?
Let's consider $f(n) = \frac{7n}5$. Whatever “computable” means, this function should have that property. For example, if we put in $n=2$ the computer can print out ...
18
votes
Accepted
How to show analytically that $x^4-4x^3-2x^2+16x+24=0$ has no solutions in $\mathbb{R}$?
$$f(x)=x^4-4x^3-2x^2+16x+24=(x^2-2x-4)^2+2x^2+8>0.\tag{1}$$
My motivation: the terms $-4x^3$ and $16x$ are annoying since they can be either positive or negative, so I wanted to put them in a ...
- 8,589
16
votes
Why isn't the Weierstrass function $\sum_{n=0}^\infty a^n \cos(b^n\pi x)$ differentiable?
It's not that it is an infinite sum, any analytic function can be written as an infinite sum by definition and is differentiable.
The issue is that the function has the nature of a fractal, at ...
- 14.9k
14
votes
Why isn't the Weierstrass function $\sum_{n=0}^\infty a^n \cos(b^n\pi x)$ differentiable?
In order to calculate the derivative of a function at a given point (e.g. $x=0$), you just need to keep zooming on this point until the curve gets smooth:
Oh, wait...
- 3,206
13
votes
Accepted
How can $f(x) = x^2$ be continuous?
You're mixing up a lot of concepts there, and that's part of what's confusing you. To give you some direction on things:
The size of sets
For finite sets, it's easy to say that their size is the ...
- 14.7k
13
votes
Accepted
When $f(z) = 1/z$ and $g(z) = 1-z$, why is $f \circ g \circ f \circ g \circ f \circ g (z) = g \circ f \circ g \circ f \circ g \circ f (z) = z$?
The functions $f(z) = 1/z$ and $g(z) = 1 - z$ are both linear fractional transformations; that is, they have the form $\frac{az + b}{cz + d}$ for suitable complex numbers $a, b, c, d$ such that $ad - ...
- 6,230
12
votes
Finding $\displaystyle \sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n(2)^n \binom{2n}{n}}$
Clear["Global`*"]
f[n_] := (-1)^
n (HarmonicNumber[2 n] - HarmonicNumber[n])/(n*2^n*Binomial[2 n, n])
The terms of the sum converge to zero rapidly
<...
- 698
11
votes
Proving $\ddot x = w-2x+x^2$, where $w\ge0$, conserves energy
In physics courses we tend to use the following trick: multiply $\ddot x = w-2x+x^2$ by $\dot x$ to obtain
$$ \dot x \ddot x = (w-2x+x^2)\dot x \ \ \Rightarrow \ \ \frac{d}{dt}\left( \frac{1}{2}\dot x^...
- 16.4k
10
votes
Functional equation $f(px)+p=[f(x)]^2$
This is a complete answer and goes a long way to explaining why the answer to this question is much more difficult than initially made out to be. Basically, there are such functions, and not only that ...
10
votes
Accepted
Isn't my book wrongly equating $\frac{\frac{\sin^2x-\cos^2x}{\sin x\cos x}}{\frac{\sin^2x+\cos^2x}{\sin x\cos x}}$ and $-\cos2x$?
$\tan x $ is not defined when $\cos x =0$ and $\cot x $ is not defined when $\sin x =0$. So it is implicitly assumed the domain excludes points where $\sin x \cos x=0$.
- 303k
10
votes
Accepted
Why is the range a larger set than the domain?
Perhaps I'm understanding your question differently than some of the other commenters, but I'll point out that saying you have a function $g: D \to \mathbb{R}$ does NOT mean every element of $\mathbb{...
- 1,788
10
votes
Accepted
Relation of Hamel basis with the equation $f(x + y) = f(x) + f(y)$?
The connection is the following one. A linear function from $\mathbb{R}$ to $\mathbb{R}$, where $\mathbb{R}$ is considered as a vector space over itself, is a function $f$ that satisfies the following ...
- 4,944
10
votes
Accepted
Will $h(x)=\frac{ax+b}{cx+d}$, where $c\neq0$, always satisfy $h(h\cdots(h(x))\cdots)=x$ in a finite number of iterations?
is it always possible that the expression of form $h(x)$ = $\dfrac{ax+b}{cx+d}$ where $c \neq 0$ will satisfy $h(h\cdots (h(x))\cdots) = x$
Let's use the following notation: For a 2×2 matrix
$$M=\...
- 8,099
9
votes
Are there non-surjective functions in ZFC theory?
Surjectivity is a relationship between a function and a set. It is an extrinsic property of a function, when the function is defined as a set of ordered pairs. Not an intrinsic one like injectivity is....
- 373k
8
votes
Accepted
$f(x^2)=f(x)+f(-x)$
There are many solutions. We need $f(0)=0$. From there, define $f(x)$ however you like for positive values of $x$. Finally, for $x<0$, let $f(x) = f(x^2) - f(-x)$.
Example: $$f(x) = \begin{cases}x^...
- 114k
8
votes
Accepted
If $\frac{f(x^2)}{f(x)}=1+x+x^2+\ldots+x^7$ then what is $f(x)$?
Simply use
$$x^n+1=\frac{x^{2n}-1}{x^n-1}$$
for each factor of your decomposition
$$(x+1)(x^2+1)(x^4+1).$$
- 5,544
8
votes
Why isn't the Weierstrass function $\sum_{n=0}^\infty a^n \cos(b^n\pi x)$ differentiable?
While others have given answers saying that a pointwise limit of differentiable functions needn’t be differentiable, here is a simple example explaining why that is true. One can find a limit of ...
- 3,824
8
votes
Accepted
In the category of Lie algebras are mono-/epimorphisms precisely the injective/surjective morphisms?
The other answer deals with monomorphisms, so let's look at epimorphisms. As pointed out in comments, there is a (1970?) preprint of G. Bergman's, Epimorphisms of Lie Algebras, available online, which ...
- 21.6k
8
votes
Accepted
Composing the empty function with itself
Notice how there really only is one function $\varnothing\to\varnothing$. Thus all your functions, i.e. $f$, $g$, $f\circ f$ and $\operatorname{id}_\varnothing$, are one and the same. The reason ...
- 3,501
8
votes
Does there exist a function which converts exponentiation into addition?
While your question as stated has been answered, I'd like to give a satisfying answer to a slight alteration of your question. To avoid the issue of commutivity, we can instead look at commutative ...
- 3,885
8
votes
Calculus: Difference between functions and "equations" from a theoretical perspective
It depends on the context what you mean with
$$f(x,y) = x^2+y^2 \tag 1$$
What it could be is the definition of a function $f:\Bbb R^2\to\Bbb R$, or over any other domains where concepts like squaring ...
- 8,099
8
votes
How do we know that mathematics is independent of the definition of the cartesian product?
One way of doing this is to introduce new notation $a, b \mapsto (a, b)$ and the axiom $(a, b) = (c, d) \to (a = c \land b = d)$ (as well as forms of replacement and separation which can have $(a, b)$ ...
- 22.5k
8
votes
Accepted
Does a function $f(x)$ exist such that $f(x+1)-f(x)=\frac{1}{x}$. And if so what is it.
Yes. Let $L(z)$ denote the logarithm of the gamma function.
Then $L(x + 1) - L(x) = \log(x)$. So the derivative $L'(x)$ (i.e. $\Gamma'(x) / \Gamma(x)$) satisfies the desired functional equation.
- 25.7k
8
votes
Find the function $f(x)$ when $f(f(x))=1-x$, for $x\in [0,1]$
From $f(f(x)) =1-x$ follows
$$1-f(x)=f(f(f(x)))=f(1-x) .$$
Hence,
$$f(x) +f(1-x) =1$$
for all $x\in[0, 1]$.
- 2,032
7
votes
Accepted
Continuous surjective function from closed interval to itself that fix only the endpoints
You have already found the simple example $f(x)=x^2$ on the interval $[0,1]$. This example translates to every other interval $[a,b]$, where $a<b$. Simply take
$$g(x)=(b-a)\left(\frac{x-a}{b-a}\...
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